# Joint probability distribution of some Gumbel differences

I would like your help to double check my derivations below involving the joint probability distribution of some Gumbel differences.

Consider $$K$$ i.i.d. random variables $$\epsilon_1,...,\epsilon_K$$, where $$\epsilon_1$$ is distributed as a Gumbel with scale $$\beta$$ and location $$\mu$$. Hence the pdf of $$\epsilon_1$$ is $$f(x)=\frac{1}{\beta}\exp\Big(-\Big(\frac{x-\mu}{\beta}\Big)\Big)\times \exp\Big(-\exp\Big(-\Big(\frac{x-\mu}{\beta}\Big)\Big)\Big)$$ and the cdf of $$\epsilon_1$$ is $$F(x)=\exp\Big(-\exp\Big(-\Big(\frac{x-\mu}{\beta}\Big)\Big)\Big)$$

Question: Let $$\alpha_1,...\alpha_K$$ be some known real numbers. Derive $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)$$

My attempt (could you check it?)

Step 1: $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1|\epsilon_1)=\Pi_{k\neq 1} Prob(\epsilon_k\leq \alpha_1-\alpha_k+\epsilon_1 |\epsilon_1)=$$ $$\exp\Big(-\exp\Big(-\Big(\frac{\epsilon_1+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big)$$

Step 2: $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{-\infty}^{\infty}\Pi_{k\neq 1}\exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big) \times \frac{1}{\beta}\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\times \exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\Big)dt$$

Step 3: Notice that for $$k=1$$ $$\exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu}{\beta}\Big)\Big)\Big)=\exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\Big)$$

Step 4: apply step 3 to step 2 $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{-\infty}^{\infty}\Pi_{k} \exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big) \times \frac{1}{\beta}\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)dt$$ Step 5: Notice that $$\Pi_{k} \exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big)=\exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big) \underbrace{\sum_{k}\exp\Big(-\Big(\frac{\alpha_1-\alpha_k}{\beta}\Big)\Big)}_{\equiv Q}\Big)$$

Step 6: apply step 5 to step 4 $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{-\infty}^{\infty}\exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\times Q\Big) \times \frac{1}{\beta}\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)dt$$

Step 7: change variables $$\omega\equiv \exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)$$ so that $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{0}^{\infty}\exp(-\omega Q)\times \frac{1}{\beta}\omega \frac{1}{\omega} d\omega=\frac{1}{\beta}\times \frac{1}{Q}$$ Hence $$Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\frac{1}{\beta \times \sum_{k}\exp\Big(-\Big(\frac{\alpha_1-\alpha_k}{\beta}\Big)\Big)}=\frac{\exp(\frac{\alpha_1}{\beta})}{\beta \sum_{k}\exp\Big(\frac{\alpha_k}{\beta}\Big)\Big)}$$

I'm wondering why $$\mu$$ does not play a role. I'm not even convinced by the $$\beta$$ at the denominator in the final expression.